# Tag Info

We calculate the free energy (density) for the Higgs field $\phi$ at finite temperature. In the Standard Model, this looks like $\mathcal{F}_{SM}(\phi,T) = -\frac{\pi^2}{90}g_* T^4+V_{SM}(\phi, T) \ ,$ where $g_*$ is the number of degrees of freedom in the SM ($g_*=106.75$). The potential has the form $V_{SM}(\phi,T) = D(T^2-T_0^2)\phi^2 - ET\phi^3+\frac{... 7 I don't think that Wick rotated time$\tau$is periodic by itself. But it turns out that thermal averages of operators are periodic with respect to the variable$\tau$. Consider a generic time dependent operator$\hat{A}(\tau)$with the standard time evolution expansion$\hat{A}(\tau) = e^{\hat{H}\tau} \hat{A}(0) e^{-\hat{H}\tau}$and consider its thermal ... 5 An enclosed cavity near thermal equilibrium at some temperature will be filled with blackbody radiation, which has a well-defined spectrum that depends only (!) on the temperature of the cavity. If you poke a small hole in the cavity, some of those photons will leak out of the hole as thermal radiation. For instance, the pupils of your eyes are dark ... 4 There are two particle candidates for dark matter, and both are still in the realm of hypothesis. These particles are candidates exactly because they cannot decay to something lighter, they are stable. Neutrinos, which also cannot decay to something lighter, has such a small mass that they cannot model the way dark matter is attracted to gravitational ... 4 If you make the substitution: $$\frac{\mathrm{i}t}{\hbar}\rightarrow\frac{1}{k_{\mathrm{B}}T} = \beta,$$ the quantum evolution operator becomes: $$\mathrm{e}^{-\mathrm{i}\frac{\hat{H}t}{\hbar}} \rightarrow \mathrm{e}^{-\beta \hat{H}},$$ which looks familiar in the context of thermal statistical field theory since the partition function$Z$is given by: $$... 4 Without temperature, if you want to know the expectation value of an operator \langle A\rangle given a field state |\psi\rangle, you just compute:$$ \langle A\rangle = \langle \psi|A|\psi\rangle . $$Thermal field theory (or just thermal quantum statistical mechanics) assumes that the observables are obtained from a system whose state is a thermal ... 3 You're using different boundary conditions around the thermal circle, so there's no reason to expect the results to be related. When you use periodic boundary conditions around the thermal circle, it's equivalent to inserting a factor (-1)^F in the trace$$Z = {\rm Tr\ } (-1)^F e^{-\beta H}$$and because of supersymmetry, only the groundstates can ... 3 That trace is nothing but the sum of the non-vanishing \rho eigenvalues to the 1/r power and taking their multiplicities into account. Since those eigenvalues \lambda belong to (0,1], then \lambda^{1/r}\to 1 as r\to +\infty. In summary tr(\rho^{1/r}) tends to the number of non-vanishing eigenvalues of \rho taking multiplicities into account. In ... 3 Thermal expansion from an atomistic perspective: The energetic potential between two atoms can be approximated by two exponential functions, one for the attractive force between the atoms, one for the repulsive force. The superposition of these two force fields has a minimum at a certain distance. Examples for such empirical potentials are Stillinger-Weber, ... 3 Roughly speaking solid matter is on a lattice form, A three-dimensional lattice filled with two molecules A and B, here shown as black and white spheres. The molecules fit like LEGO , the forces tying them together are mainly the spill over electric field forces , attractive and repulsive forming the patterns of the lattice. In a single crystal one ... 3 Let's define T_{EW} the temperature where the coefficient m^2_H(T) of the operator H^2 in the SM lagrangian vanishes:$$ m_H^2(T=T_{EW})=0\,. $$For T>T_{EW} the Higgs vev is vanishing, the EW symmetry in unbroken, and the elementary particles are massless. For T<T_{EW} the the vev is non-vanishing, v_T\propto -m_H^2/\lambda\neq 0, the EW ... 3 That is the same situation as in QFT, as soon as you drop Poincaré invariance. I mean in curved spacetime. There you may have uncountably many inequivalent representations of the same algebra of observables just by varying some continuous parameter (curvatures). If you put all these reps in orthogonal sectors in a common Hilbert space, it must be non.... 3 Chaos is common property of systems in general, not only thermal ones - in order to avoid chaos, you need to pick a system by hand, with a simple enough Hamiltonian, which you can't for the typical "thermal system". In thermodynamics, though, the connection is deeper: thermal systems (in equilibrium) "can be thought of as 'moving through phase space' ... 2 Collisions of a state with other particles, present at finite density, influence the life-time of the state as energy can be transferred to those other particles during inelastic collisions (i.e. they can change state). This change in life-time is related to a change in the width via the uncertainty relation and gives the thermal width. There is a physical ... 2 Finite temperature Feynman rules are simply zero temperature Feynman rules for Euclidean (t\to i\tau) QFT in periodic imaginary time. So instead of continuous values for the momenta, you will have a discrete spectrum for the timelike moments (such as in the infinite potential well in basic quantum mechanics). It's called the Matsubara formalism, if you ... 2 1) What you call \rho should really be called \epsilon (this is the energy density, not the particle density). 2) The thermodynamic variables \epsilon and P are expectation values of certain operators in a thermal ensemble. You should not confuse equations for the operators with equations for thermodynamic quantities. 3) The operator that ... 2 There seems to be some confusion in your question between thermal fluctuations and quantum fluctuations, so I will try to address both of them in my answer. Spontaneous symmetry breaking occurs when the ground state of the Hamiltonian does not exhibit the full symmetries of the Hamiltonian. In other words, there are multiple degenerate configurations with ... 2 The point is that to achieve the sum over Matsubara frequencies$$\sum_{n} g(i\omega_n)$$we can use a contour integral$$\oint_C g(z) f(z)$$with the contour described in fig 1 here, so long as we choose an f(z) with simple poles exactly at the Matsubara frequencies \omega_n. This determines f(z) to be proportional to the Bose-Einstein distribution. ... 2 A Euclidean correlation function may be interpreted as a Lorentzian expectation value by "cutting" the path integral and continuing the time coordinate. Let me review how this procedure relates Euclidean correlation functions on a closed manifold M\times S^1_\beta to thermal expectation values in a Lorentzian quantum field theory on M \times \mathbb{R}... 2 Two remarks: Essentially every function can be regarded as a distribution, although the converse is not true. In this sense, you can indeed regard (e^{\beta |p_0|} - 1)^{-1} as a distributions. Distributions of this kind are known as regular distributions. Non-regular distributions are known as singular distributions. It is not true that you cannot ... 2 First write \rho, y in the eigenbasis of the Hamiltonian,$$ \rho = \sum_n p_n |n\rangle \langle n| \\ y = \sum_n q_n |n\rangle \langle n| $$with q_n^4 = p_n, in particular 0\le q_n \le 1. In this basis the inequality becomes$$ \sum_{n,m} |V_{n,m}|^2 q_n^{1+\eta} q_m^{3-\eta} \le \sum_{n,m} |V_{n,m}|^2 q_n q_m^2 $$At this point the ... 2 Note that at finite temperature we don't have scattering amplitudes in quite the same way that we do at T=0, because the notion of an asymptotic state does not quite exist. The natural observables are the partition function, and correlation functions of gauge invariant operators. But let's ignore this issue, and look at ee scattering. At T=0 the ... 2 Thomas gave you a good answer, but here’s an additional perspective … At finite temperature, the so-called vacuum isn’t all that empty. It is suffused by blackbody radiation and a miasma of electron-positron pairs. If a diagram of interest contains a on-shell boson propagator, and the not-so-empty environment contains a photon line, you must now add the ... 2 Constructing a macroscopic relativistic quantum field theory is done in exactly the same way all such quantum theories are. You assume the system is in equilibrium, i.e. at some temperature T. The partition function for a quantum system is given by$$\mathcal{Z}=\text{tr}\big(e^{-\beta H}\big)$$You use this to calculate all sorts of statistically ... 1 For a Yukawa interaction of the type$$\mathscr{H}_{int}=-\lambda \bar{\psi}_B\psi_B\phi_A,\tag{1}$$where \phi is a scalar field of mass M_A and \psi is a fermionic field of mass M_B (with M_A>2M_B), is responsible for a decay of the form$$A\to B+\bar{B}.\tag{2}$$The exact zero-temperature decay rate at the tree-level is given by$$\Gamma_{\... 1 I am not going to go into the black hole stuff, but thermalization is never loss of matter and/or probability, but - so to say - only a "rearrangement" of density matrix eigenstates and corresponding probabilities. In other words, it is an evolution from an arbitrary initial state$\rho$, which may be pure ($\;\rho = \rho^2 = |\Psi\rangle\langle \Psi |\;$) ... 1 The effective action is defined so that$\frac{\delta \Gamma \left[ \phi \right]}{\delta \phi^i \left( x \right)}=J_{\phi}^i \left( x \right)$, where$J$is a classical current, so we see that this definition coincides with the usual definition of the action in classical field theory. Moreover, the procedure of evaluation of this action involves evaluation ... 1 Concerning OP's explicit example $$\frac{\delta(p^2 + m^2 )}{e^{\beta |p_0|} - 1} , \qquad\beta\neq 0. \tag{D}$$ The massive case$m\neq 0$. Then the singularity of$\frac{1}{e^{\beta |p_0|} - 1}$does not overlap with the support of$\delta(p^2 + m^2 )$, so the product distribution is mathematically well-defined. The massless case$m=0$. Then the ... 1 No. Consider for example observables of the form$\mathcal O = e^{\beta H/2}\$; the value exponentially increases as the probability exponentially decreases and you can get what's in principle an infinite sum of constant terms.